CIRCLE THEOREMS Word Problems

1. PQRS is a cyclic quadrilateral. Side PS and QR are produced to meet at a point T

outside the circle. If PQ=RT and ̂ = ̂ , prove that: ∠ = ∠

Given: PQRS is a cyclic quadrilateral. Side PS and QR P S
are produced to meet at a point T outside the R
To prove: circle. Also, PQ=RT and P̂S = R̂S T
Construction:
∠PRQ = 2∠RTS
QS is joined

Proof: Q

Statements Reasons

1. In ∆PQS and ∆RTS 1.
i) PS = RS (S) i) Corresponding chords to equal arcs, P̂S = R̂S.

ii) ∠QPS = ∠TRS (A) ii) Relation of exterior and opposite interior angles of cyclic quad.
iii) PQ = RT (S) iii) Given

2. ∆PQS ≅ ∆RTS 2. By S.A.S. axiom

3. ∠PQS = ∠RTS 3. Being corresponding sides of congruent triangles

4. ∠PQS = ∠RPS 4. Being inscribed angled standing on equal arcs

5. ∠PRQ = ∠RPS + ∠RTS 5. Relation of exterior and opposite interior angles of ∆

6. ∠PRQ = 2∠RTS 6. From statements 3, 4 and 5.

1Proved

2. Points S, T, A and R are con-cyclic such that arc ST = arc AR. If the chords SA and

TR are intersected at E, prove that: i) Area of ∆ = Area of ∆ ii) SA = TR

Given: Points S, T, A and R are con-cyclic such that S T
To prove: arc ST = arc AR, the chords SA and TR are RE
intersected at E
i) Area of ∆SET = Area of ∆ARE ii) SA = TR

Proof: A

Statements Reasons
1. In ∆SET and ∆ARE 1.

i) ∠TSE = ∠ARE (A) i) Being inscribed angle standing on same arc.
ii) ST = RA (S) ii) Being corresponding chords to given ŜT = ÂR
iii) ∠ETS = ∠EAR (A) iii) Same as reason 1(i)
2. ∆SET ≅ ∆ARE
3. Ar. of ∆SET = Ar. of ∆ARE 2. By A.S.A. axiom
4. ŜT = ÂR 3. Being congruent triangles
5. ŜT + R̂S = ÂR + R̂S 4. Given
i.e. T̂SR = ŜRA
6. TR = SA 5. By adding common R̂S in statement 1

6. Being corresponding chords to equal arcs 2Proved

3. SONG is a cyclic quadrilateral. If the bisectors of ∠ and ∠ meet the circle at

P and Q respectively, then prove that PQ is diameter of the circle. P
SONG is a cyclic quadrilateral. If the bisectors of ∠SGN O
Given: N

and ∠SON meet the circle at P and Q respectively

To prove: PQ is diameter of the circle S

Construction: QG is joined Reasons G
Proof: Q

Statements

1. ∠SON + ∠SGN = 180° 1. Being opposite angles of cyclic quadrilateral

2. ∠SOQ = 1 ∠SON and ∠SGP = 1 ∠SGN 2. Given OQ and GP are bisector of ∠SON and ∠SGN
22
3. ∠SOQ + ∠SGP = 90° 3. From above statements and solving

4. ∠SOQ = ∠SGQ 4. Being inscribed angles standing on same arc

5. ∠SGQ + ∠SGP = 90° 5. From statements 3 and 4
i.e. ∠PGQ = 90° 6. Being ∠PGQ = 90° from statement 5.

6. PQ is diameter of the circle

Proved

3

4. In a cyclic quadrilateral NEST, NS = TE, prove that: i) NT = ES ii) NE//ST

iii) Ar.∆ = Ar.∆ N

Given: In a cyclic quadrilateral NEST, NS = TE T

To prove: i) NT = ES ii) NE//ST iii) Ar.∆SEN = Ar.∆TEN E

Construction: NE and ST is joined

Proof: S

Statements Reasons
1. N̂ES = ÊNT 1. Being corresponding arcs to given chords NS = TE
2. N̂ES − N̂E = ÊNT − N̂E 2. Subtracting common arc from statement 1

i.e. ÊS = N̂T 3. Being ÊS = N̂T from statement 2
3. ES = NT and NE//ST 4.
4. In ∆SEN and ∆TEN
i) Being corresponding angles to ÊS = N̂T from st. 2
i) ∠SNE = ∠TEN (A) ii) Given
ii) NS = TE (S) iii) Being inscribed angles standing same arc
iii) ∠ESN = ∠ETN (A) 5. By A.S.A. axiom
5. ∆SEN ≅ ∆TEN 6. Being congruent triangles.
6. Ar.∆SEN = Ar.∆TEN
Proved

4

5. ABCD is a cyclic quadrilateral. A side CD is produced to the point P. BD and AC are

diagonals. If AB = AC, prove that AD is bisector of ∠ .

Given: ABCD is a cyclic quadrilateral, a side CD is B A
produced to the point P, BD and AC are P
diagonals, AB = AC

To prove: AD is bisector of ∠BDP D

Proof: C

Statements Reasons

1. ∠ADB = ∠ACB 1. Being inscribed angles standing on same arc
2. ∠ACB = ∠ABC 2. Being angles opposite to given AB = AC
3. ∠ABC = ∠ADP 3. Being exterior and opposite interior angles of cyclic quad.
4. ∠ADB = ∠ADP 4. From above statements
5. AD is bisector of ∠BDP 5. Being ∠ADB = ∠ADP from statement 4

Proved

5

6. O is centre of the circle. If two chords DE and FG intersect at point at point H and

points D, E, F and G are joined with centre O, then prove that: ∠ + ∠ =

∠ O is centre of the circle, two chords DE and FG D G
Given: intersect at point at point H and points D, E, F O

and G are joined with centre O H
FE
To prove: ∠DOF + ∠EOG = 2∠EHG

Construction: FE is joined

Proof:

Statements Reasons

1. ∠EHG = ∠EFG + ∠DEF 1. Being exterior and opposite interior angles of a triangle

2. ∠EFG = 1 ∠EOG and ∠DEF = 1 ∠DOF 2. Being inscribed and central angles on same arcs
22
3. 11 3. From above statements
∠EHG = 2 ∠EOG + 2 ∠DOF
4. 2∠EHG = ∠EOG + ∠DOF 4. Solving statement 3

Proved

6

7. In a cyclic quadrilateral ABFE, C and D are two points on BF and AE respectively

such that EF//CD. Prove that ABCD is a cyclic quadrilateral. A
D
Given: In a cyclic quadrilateral ABFE, C and D are two B
C
points on BF and AE respectively such that

EF//CD

To prove: ABCD is a cyclic quadrilateral F

Proof: E

Statements Reasons

1. ∠BAD + ∠CFE = 180° 1. Being opposite angles of cyclic quadrilateral ABFE
2. ∠CFE = ∠BCD 2. Being corresponding angles, given EF//CD
3. ∠BAD + ∠BCD = 180° 3. From above statements
4. ABCD is a cyclic quadrilateral 4. Being opposite angles of quad. ABCD supplementary

Proved

7

8. The sides AD and BC of cyclic quadrilateral ABCD is produced to E and F

respectively in such a way that, AB//EF. Prove that CDEF is a cyclic quadrilateral.

Given: The sides AD and BC of cyclic quadrilateral

ABCD is produced to E and F respectively in A DE
such a way that, AB//EF

To prove: CDEF is a cyclic quadrilateral

Proof: B CF

Statements Reasons

1. ∠BAD + ∠DEF = 180° 1. Being co-interior angles, given AB//EF

2. ∠BAD = ∠DCF 2. Being exterior and opposite interior angles of cyclic quad

3. ∠DCF + ∠DEF = 180° 3. From above statements

4. CDEF is a cyclic quadrilateral 4. Being opposite angles of quadrilateral supplementary

Proved

8

9. In a cyclic hexagon ABCDXY the diagonals AD, BX and CY are drawn. If AD//BC

then prove that ∠ = ∠ . AB

Given: In a cyclic hexagon ABCDXY the diagonals AD,

To prove: BX and CY are drawn, AD//BC YC
∠AYC = ∠BXD

Proof: XD
Reasons
Statements 1. Being arcs intersected between the parallel lines AD//BC
1. ÂB = ĈD 2. Adding common arc ÂB in statement 1
2. ÂB + B̂C = ĈD + B̂C
3. Being inscribed angles standing on equal arcs from st. 2
i.e. ÂBC = B̂CD Proved
3. ∠AYC = ∠BXD

9

10. In a cyclic quadrilateral ABCE, AB//EC. If D be any point on EC such that DA//BC

is drawn, prove that AE = AD. C

Given: In a cyclic quadrilateral ABCE, AB//EC, D be any point DE

on EC such that DA//BC is drawn

To prove: AE = AD

Proof: BA

Statements Reasons
1. ∠ADE + ∠ADC = 180° 1. Being linear pair
2. ∠ADC = ∠ABC 2. Being opposite angles of parallelogram ABCD
3. ∠ADE + ∠ABC = 180° 3. From above statements
4. ∠ABC + ∠AED = 180° 4. Being opposite angles of cyclic quadrilateral ABCE
5. ∠ADE + ∠ABC = ∠ABC + ∠AED 5. From statements 3 and 4
6. ∠ADE = ∠AED 6. Cancelling common angle ∠ABC from statement 5
7. AE = AD 7. Being opposite sides to equal angles of ∆AED

Proved

10

11. In a circle, chords RP and SQ are produced to meet at O. If PQ//RS, prove that OP

= OQ. In a circle, chords RP and SQ are produced to R
Given: meet at O, PQ//RS
OP = OQ P
To prove: O

Q

Proof: S

Statements Reasons
1. ∠PRS = ∠OPQ 1. Being corresponding angles, given PQ//RS
2. ∠PRS = ∠OQP 2. Being exterior and opposite interior angles of cyclic quad. PQSR
3. ∠OPQ = ∠OQP 3. From above statements
4. OP = OQ 4. Being opposite sides to equal angles of ∆OPQ

Proved

11

12. In a semi-circle, AOB is diameter. C and D are two points at the circumference

such that arc DC = arc BC, then prove that AD//OC. DC

Given: In a semi-circle, AOB is diameter. C and D are two

points at the circumference such that arc DC = arc BC

To prove: AD//OC AO B
Proof:

Statements Reasons

1. ∠BAD ≗ 1 B̂CD 1. Being inscribed angles and their corresponding arcs
2
2. 1 2. Given D̂C = B̂C
B̂C = 2 B̂CD
3. From above statements
3. ∠BAD ≗ B̂C 4. Being central angle and its corresponding arc
5. From statement 3 and 4
4. ∠BOC ≗ B̂C 6. Being corresponding angles equal from statement 5.

5. ∠BAD = ∠BOC

6. AD//OC

Proved

12

13. In a circle, chords PQ and RS intersect within the circle at point X. Prove that:

∠ ≗ ( ̂ + ̂ ). S
P


Given: In a circle, chords PQ and RS intersect within

To prove: the circle at point X X Q
R
Construction: ∠PXR ≗ 1 (P̂R + Q̂S)
Proof:
2

RQ is joined

Statements Reasons

1. ∠PXR = ∠PQR + ∠QRS 1. Being exterior and opposite interior angles of a triangle
2. Being inscribed angles and their corresponding arcs
2. ∠PQR ≗ 1 P̂R and ∠QRS ≗ 1 Q̂S 3. From above statements
22
3. 1 1 Proved
∠PXR ≗ 2 P̂R + 2 Q̂S

i.e. ∠PXR ≗ 1 (P̂R + Q̂S)
2

13

14. In a circle, chords CE and BD are produced to the external point A. If AB = AC,

then prove that BE = CD. C

Given: In a circle, chords CE and BD are produced to E
A
the external point A, AB = AC

To prove: BE = CD

D

Proof: B

Statements Reasons
1. In ∆ABE and ∆ACD 1.

i) ∠BAE = ∠CAD (A) i) Being common angles
ii) AB = AC (S) ii) Given
iii) ∠DBE = ∠DCE (A) iii) Being inscribed angles on same arc DE
2. ∆ABE ≅ ∆ACD 2. By A.S.A axiom
3. BE = CD 3. Being corresponding sides of congruent triangles

Proved

14

15. D, E and F are the midpoints of sides AB, AC and BC A
respectively and ⊥ then prove that DEFG is a cyclic

quadrilateral.

Given: D, E and F are the midpoints of sides AB, AC and BC E D

respectively and AG ⊥ BC

To prove: DEFG is a cyclic quadrilateral C FG B
Proof:

Statements Reasons

1. AD = BD = GD 1. GD being median drawn to the hypotenuse of rt. angled ∆
2. ∠DBG = ∠DGB 2. Being GD = BD from statement 1
3. BFED is a parallelogram 3. Being EF//AB and ED//CB, (lines joining midpoints of any two sides of a

triangle is parallel to third side).

4. ∠DBG = ∠FED 4. Being opposite angles of parallelogram BFED
5. ∠DGB = ∠FED 5. From statement 2 and 4
6. DEFG is a cyclic quad. 6. Being exterior and opposite interior angles of quad. equal

Proved

15

16. In a circle with centre O, the chord BC is intersected by a radius OA. If ̂ = ̂

and BC is an angle bisector of ∠ , prove that ∆ is an equilateral triangle.

Given: In a circle with centre O, the chord BC is intersected B A
by a radius OA, ÂC = ÂB and BC is an angle bisector

To prove: of ∠ABO C
∆ABO is an equilateral triangle O

Proof: Reasons
1. Being inscribed and central angles standing on same arc
Statements 2. Given BC is an angle bisector of ∠ABO
1. ∠AOB = 2∠ACB 3. Being inscribed angles standing on equal arcs, given ÂC = ÂB
2. ∠ABO = 2∠ABC
3. ∠ACB = ∠ABC

4. ∠AOB = ∠ABO 4. From above statements

5. AB = OA = OB 5. Being ∠AOB = ∠ABO and OA = OB = radius

6. ∆ABO is an equilateral 6. Being all sides equal
triangle

Proved

16

17. In a circle with centre O, PR is diameter and Q is a point n circumference of the

circle such that ̂ : ̂ = : . Prove that ∆ is an equilateral triangle. P

Given: In a circle with centre O, PR is diameter and Q Q

is a point n circumference of the circle such

To prove: that Q̂R: P̂Q = 2: 1 O
∆POQ is an equilateral triangle

Proof: R

Statements Reasons

1. ∠PQR = 90° 1. Being inscribed in semi-circle

2. ∠QPR + ∠QRP = 90° 2. Being acute angles of right angled ∆PQR

3. ∠QPR = 2∠QRP 3. Being inscribed angles standing on given arcs, Q̂R: P̂Q = 2: 1

4. ∠QPR = 60° & ∠QRP = 30° 4. From statements 2 and 3

5. ∠POQ = 2∠QRP = 60° 5. Being inscribed and central angles standing on P̂Q

6. ∠QPR = ∠POQ = ∠PQO 6. From statements 4 and 5 and being OP = OQ = r

7. ∆POQ is an equilateral 7. Being each interior angles equal 60°
triangle

Proved

17

18. ABNM is a cyclic quadrilateral. C and D are two points on the chord AB. If AC = BD

and ̂ = ̂ , prove that ∠ = ∠ and . ∆ = . ∆ . AC

Given: ABNM is a cyclic quadrilateral, C and D are two points DB

on the chord AB, AC = BD and ÂM = B̂N

To prove: ∠ACM = ∠BDN and Ar. ∆ACM = Ar. ∆BDN M

Construction: AM and BN are joined

Proof: N

Statements Reasons

1. ÂM = B̂N 1. Given
2. By adding M̂N in statement 1 and solving
2. ÂM + M̂N = B̂N + M̂N
i.e. ÂMN = B̂NM 3.
i) Being corresponding chords to given ÂM = B̂N.
3. In ∆MAC and ∆NBD ii) Being inscribed angles standing on equal arcs from statement 2
i) AM = BN (S) iii) Given
ii) ∠MAC = ∠NBD (A)
iii) AC = BD (A)

4. ∆MAC ≅ ∆NBD 4. By S.A.S. axiom
5. ∠ACM = ∠BDN 5. Being corresponding angles of congruent triangles
6. Ar. ∆ACM = Ar. ∆BDN 6. Being congruent triangles

Proved

18

19. In a circle, chords MN and RS when produced, intersect externally at point X.

Prove that: ∠ ≗ ̂ − ̂ . M

Given: In a circle, chords MN and RS when produced, N

intersect externally at point X X

To prove: 2∠MXR ≗ M̂R − N̂S S
R
Construction: RN is joined

Proof:

Statements Reasons

1. ∠MNR ≗ 1 M̂R and ∠NRS ≗ 1 N̂S 1. Being inscribed angle and its corresponding arc
22
2. ∠MXR = ∠MNR − ∠NRS 2. Being exterior and opposite interior angles of a triangle

3. ∠MXR ≗ 1 M̂R − 1 N̂S 3. From above statements
2 2 4. Solving statement 3.
4. 2∠MXR ≗ M̂R − N̂S

Proved

19

20. In a circle, two chords AB and CD intersect at right angles at X. Prove that ̂ −

̂ = ̂ − ̂ . D

Given: In a circle, two chords AB and CD intersect at

right angles at X A

To prove: ÂD − ĈA = B̂D − B̂C X

Construction: AC and AD are joined B
Proof: C

Statements Reasons

1. ∠BAD ≗ 1 B̂D and ∠CDA ≗ 1 ĈA 1. Being inscribed angle and its corresponding arc

22 2. Same as reason 1

2. ∠BAC ≗ 1 B̂C and ∠ACD ≗ 1 ÂD 3. Being sum of acute angles of right angled triangles
4. From above statements
22

3. ∠BAD + ∠CDA = ∠BAC + ∠ACD

4. 1 B̂D + 1 ĈA = 1 B̂C + 1 ÂD
2 2 2 2
5. ÂD − ĈA = B̂D − B̂C 5. Solving statement 4

Proved

20

21. In a circle having centre O, diameter AB and chord DC are produced to the

external point E. If triangle OCE is an isosceles triangle, prove that ̂ = ̂ .

Given: In a circle having centre O, diameter AB and D

chord DC are produced to the external point E, C

To prove: triangle OCE is an isosceles triangle A O BE
3B̂C = ÂD

Construction: OD is joined

Proof:

Statements Reasons

1. ∠BOC = ∠OED 1. Being base angles of isosceles triangle, given OC = CE
2. Being exterior and opposite interior angles of a triangle,
2. ∠OCD = ∠BOC + ∠OED
i.e. ∠OCD = 2∠BOC from statement 1 and solving.
3. Being OC = OD = radius and from statement 2.
3. ∠ODC = ∠OCD = 2∠BOC
4. ∠AOD = ∠ODC + ∠OED 4. Being exterior and opposite interior angles of a triangle
(From statements 1 and 3)
or, ∠AOD = 2∠BOC + ∠BOC (Solving)
⸫ ∠AOD = 3∠BOC
5. ∠AOD ≗ ÂD 5. Being central angle and its corresponding arc
6. ∠BOC ≗ B̂C
7. 3B̂C = ÂD 6. Same as reason 5

7. From statements 4, 5 and 6.

21Proved

22. In a circle with centre O, two chords PQ and MN are in a such a way that arc PM =

arc QN. Prove that chord PQ//chord MN. P

Given: In a circle with centre O, two chords PQ and MN Q

are in a such a way that arc PM = arc QN

To prove: chord PQ//chord MN M
Construction: MQ is joined N

Proof:

Statements Reasons

1. P̂M = Q̂N 1. Given
2. ∠PQM = ∠NMQ
3. PQ//MN 2. Being inscribed angles standing on equal arcs
3. Being alternate angles equal

Proved

22

23. In a circle, AB and CD are two chords. The inscribed angles on arcs AC and BD,

∠ and ∠ respectively are equal, prove that AB//CD. A

Given: In a circle, AB and CD are two chords, the B

inscribed angles on arcs AC and BD, ∠APC and

To prove: ∠BQD respectively are equal C
AB//CD D

Construction: BC is joined PQ
Proof:

Statements Reasons

1. ∠ABC = ∠APC 1. Being inscribed angles standing on same arc AC
2. ∠APC = ∠BQD 2. Given
3. ∠BQD = ∠BCD 3. Being inscribed angles standing on same arc BD
4. ∠ABC = ∠BCD 4. From above statements
5. AB//CD 5. Being opposite sides to equal angles of a triangle

Proved

23

24. XY is a line parallel to the base BC of an isosceles triangle ABC and formed by

joining the midpoints of equal sides AB at X and AC at Y. Prove that the points X,

B, C and Y are con-cyclic. A

Given: XY is a line parallel to the base BC of an isosceles triangle

ABC and formed by joining the midpoints of equal sides X Y
AB at X and AC at Y

To prove: points X, B, C and Y are con-cyclic BC
Proof:

Statements Reasons

1. ∠ABC = ∠ACB 1. Being base angles if an isosceles triangle
2. ∠ACB = ∠AYX 2. Being corresponding angles, given XY//BC

3. ∠ABC = ∠AYX 3. From above statements
4. X, B, C and Y are con-cyclic 4. Being exterior and opposite interior angles of quad.

XBCY equal from statement 3

Proved

24

25. In ∆ , ⊥ and ⊥ . If E and D are joined then prove that ∠ = ∠ .
A
Given: In ∆ABC, AD ⊥ BC and CE ⊥ AB, E and D are

joined

To prove: ∠BDE = ∠BAC E

Proof: B DC

Statements Reasons
1. AEDC is cyclic quad. 1. Being inscribed angles standing on arc AC 90º, given AD ⊥ BC and CE ⊥ AB
2. ∠ = ∠ 2. Being exterior and opposite interior angles of cyclic quadrilateral

Proved

25

26. PQRS is a parallelogram. The circle making PS as a chord cuts PQ at M and RS at

N. Prove that ∠ = ∠ . S NR

Given: PQRS is a parallelogram, the circle making PS as

a chord cuts PQ at M and RS at N

To prove: ∠MNS = ∠PQR P MQ

Constructin: MN is joined

Proof:

Statements Reasons

1. ∠QPS + ∠MNS = 180° 1. Being opposite angles of cyclic quadrilateral
2. ∠QPS + ∠PQR = 180° 2. Being co-interior angles of parallelogram
3. ∠QPS + ∠MNS = ∠QPS + ∠PQR 3. From above statements
4. ∠MNS = ∠PQR 4. Solving statement 3

Proved

26

27. ABC is a triangle with AB = AC. Also a circle passing through B and C intersects AB

and AC at D and E respectively. Prove that AD = AE. B

Given: ABC is a triangle with AB = AC, also a circle D

passing through B and C intersects AB and A

To prove: AC at D and E respectively E
AD = AE

Proof: C

Statements Reasons

1. ∠DBC = ∠ECB 1. Being angles, opposite to given sides, AB = AC

2. ∠DBC = ∠AED 2. Being exterior and opposite interior angles of cyclic quad.

3. ∠ECB = ∠ADE 3. Same as reason 2.

4. ∠AED = ∠ADE 4. From above statements

5. AD = AE 5. Being opposite sides to equal angles of a triangle

Proved

27

28. ABC is a triangle with AB = AC. Also a circle passing through B and C intersects AB

and AC at D and E respectively. Prove that BC//DE. B

Given: ABC is a triangle with AB = AC, also a circle D

passing through B and C intersects AB and A

To prove: AC at D and E respectively E
BC//DE

Proof: C

Statements Reasons

1. ∠DBC = ∠BCE 1. Being base angles opposite to given sides AB = AC

2. ∠DBC = ∠DEA 2. Being exterior and opposite interior angles of cyclic quad

3. ∠BCE = ∠DEA 3. From above statements

4. BC//DE 4. Being corresponding angles equal

Proved

28

29. In ORS, a circle passing through R and S intersects OR and OS at the points P and

Q respectively. If PQ//RS then prove that OP = OQ. R P

Given: In ORS, a circle passing through R and S O

intersects OR and OS at the points P and Q

To prove: respectively, PQ//RS Q
OP = OQ

Proof: S

Statements Reasons

1. ∠PRS = ∠OPQ 1. Being corresponding angles, given PQ//RS
2. Being exterior and opposite interior angles of cyclic quad.
2. ∠PRS = ∠OQP 3. From above statements
3. ∠OPQ = ∠OQP 4. Being opposite sides to equal angles of a triangle
4. OP = OQ
Proved

29

30. PQR is an isosceles triangle with PQ = PR. A circle is drawn with PQ as a diameter
R
and cuts the base QR at a point S. Prove that QS = RS.

Given: PQR is an isosceles triangle with PQ = PR, a

circle is drawn with PQ as a diameter and S

cuts the base QR at a point S

To prove: QS = RS P
QO
Construction: PS is joined

Proof: Reasons

Statements 1. Being inscribed angle on semi-circle, ∠PSQ = 90°
1. PS ⊥ QR 2.
2. In ∆PSQ and ∆PSR
i) Being PS ⊥ QR, from statement 1.
i) ∠PSQ = ∠PSR (R) ii) Given
ii) PQ = PR (H) iii) Being common sides
iii) PS = PS (S)
3. ∆ABC ≅ ∆DBC 3. By R.H.S. axiom
4. QS = RS
4. Being corresponding sides of congruent triangles

Proved

30

31. AOB is a diameter of a circle ABC. Through C, a straight line AD is drawn to cut the

circle at the point of circumference C such that AC = CD and BD is joined. Prove

that ∆ is an isosceles. D

Given: AOB is a diameter of a circle ABC, through C,

a straight line AD is drawn to cut the circle at C
the point of circumference C such that AC =

To prove: CD and BD is joined B
∆BAD is an isosceles AO

Construction: BC is joined

Proof:

Statements Reasons

1. BC ⊥ AD 1. Being inscribed angle on semi-circle, ∠ACB = 90° 31Proved
2.
2. In ∆ABC and ∆DBC
i) AC = CD (S) i) Given
ii) ∠ACB = ∠DCB (A) ii) Being BC ⊥ AD, from statement 1.
iii) BC = BC (S) iii) Being common sides

3. ∆ABC ≅ ∆DBC 3. By S.A.S. axiom

4. AB = BD 4. Being corresponding sides of congruent triangles
5. Being two sides of triangle equal from statement 4
5. ∆BAD is an isosceles

32. In a circle with centre O, two chords MN and XY are produced to meet at A. Prove

that ∠ − ∠ = ∠ . X

Given: In a circle with centre O, two chords MN and Y
XY are produced to meet at A
A
To prove: ∠XOM − ∠YON = 2∠XAM O
MN
Construction: XN is joined

Proof:

Statements Reasons

1. 1 1. Being inscribed and central angles standing on same arc
∠XNM = 2 ∠XOM
2. Same as reason 1
2. 1
∠YXN = 2 ∠YON 3. Being relation between exterior and opposite interior angles of ∆AXN
3. ∠XAM = ∠XNM − ∠YXN 4. From above statements

4. 11 5. Solving statement 4
∠XAM = 2 ∠XOM − 2 ∠YON
5. 2∠XAM = ∠XOM − ∠YON

Proved

32

33. PQR is an equilateral triangle inscribed in a circle. If PS is diameter of the circle, prove
R
that PS = 2SR.

Given: PQR is an equilateral triangle inscribed in a circle, O is S

the centre and PS is diameter of the circle

To prove: PS = 2SR O

Construction: OR and RS are joined PQ
Proof:

Statements Reasons

1. ∠PQR = 60° 1. Being an angle of equilateral triangle, ∆PQR
2. i) Being inscribed angles standing on same arc and from st. 1.
2. i) ∠RSO = ∠PQR = 60°
ii) ∠PRS = 90° ii) Being inscribed angles on semi-circle, given PS is diameter
iii) ∠RPS = 30° iii) Being remaining angles of triangle, ∆PRS.

3. ∠ROS = 2∠RPS = 60° 3. Being inscribed and central angles standing on same arc and from statement 2(iii).
4. OR = OS = SR 4. Being ∠ROS = ∠RSO, from statements 2(i) and 3, OR = OS = radius
5. Being diameter and radius, also from statement 4
5. PS = 2OS = 2SR
Proved

33

34. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

34

35. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

35

36. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

36

37. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

37

38. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

38

39. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

39

40. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

40

41. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

41

42. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

42

43. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

43

44. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

44

45. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

45

47. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

46

47. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

47

48. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

48

49. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

49

50. The diagonals AC and BD of a cyclic quadrilateral ABCD are intersected at point E.

From E perpendicular is drawn to AB at G. GE is produced to meet CD at F. Prove

that DF = FC.

Given: The diagonals AC and BD of a cyclic quadrilateral DFC
ABCD are intersected at point E perpendicularly,

To prove: from E perpendicular is drawn to AB at G. GE is E
Proof: produced to meet CD at F AG B
DF = FC

Statements Reasons

1. In ∆GBE and ∆DCE 1.
i) ∠GBE = ∠DCE i) Being inscribed angles standing on same arc, ÂD.
ii) ∠BGE = ∠DEC ii) Given AC ⊥ BD and GE ⊥ AB.
iii) ∠GEB = ∠CDE iii) Being remaining angles of triangles

2. ∠GEB = ∠DEF 2. Being vertically opposite angles
3. Being ∠CDE = ∠DEF, from statements 1(iii) and 2
3. DF = EF 4. Similar as above statements
5. From statements 3 and 4
4. FC = EF

5. DF = FC

Proved

50